The match of the responses to the letters gives the solution of the riddle
as; IT HAS THE EDGE
Which method can be used to find the volume of a cone?
1. The volume of a cone is; [tex]V = \mathbf{\frac{1}{3} \cdot \pi \cdot r^2 \cdot h}[/tex]
Therefore;
[tex]Volume \ of \ the \ cone \approx \frac{1}{3} \times3.14 \times 7^2 \times 14 \approx \mathbf{718 \, u^3}[/tex]
718 u³ is equivalent to S
2. Diameter of the cone, d = 24
Therefore;
Radius of the cone = 12
[tex]Volume \ of \ the \ cone \approx \frac{1}{3} \times3.14 \times 12^2 \times 28 \approx \mathbf{4220.2\, u^3}[/tex]
4220.2 u³ is equivalent to E
3. d = 11, and h = 12
Therefore;
r = 5.5
[tex]Volume \ of \ the \ cone \approx \frac{1}{3} \times 3.14 \times 5.5^2 \times 12\approx \mathbf{379.9\, u^3}[/tex]
379.9 u³ is equivalent to T
4. r = 3.2, and h = 10
Therefore;
[tex]Volume \ of \ the \ cone \approx \frac{1}{3} \times 3.14 \times 3.2^2 \times 10\approx \mathbf{107.2\, u^3}[/tex]
107.2 u³ is equivalent to G
5. r = 9, h = 15
[tex]Volume \ of \ the \ cone \approx \frac{1}{3} \times 3.14 \times 9^2 \times 15\approx \mathbf{1271.7\, u^3}[/tex]
1,271.7 u³ is equivalent to D
6. d = 4, and h = 6
Therefore;
r = 2
[tex]Volume \ of \ the \ cone \approx \frac{1}{3} \times 3.14 \times 2^2 \times 6\approx \mathbf{25.1 \, u^3}[/tex]
25.1 u³ is equivalent to A
7. r = 1.5, and h = 3
Therefore;
[tex]Volume \ of \ the \ cone \approx \frac{1}{3} \times 3.14 \times 1.5^2 \times 3\approx \mathbf{7.1\, u^3}[/tex]
7.1 u³ is equivalent to E
8. d = 30, and h = 34
Therefore;
r = 15
[tex]Volume \ of \ the \ cone \approx \frac{1}{3} \times 3.14 \times 15^2 \times 34\approx \mathbf{8,007\, u^3}[/tex]
8,007 u³ is equivalent to I
9. r = 5, and h = 8
[tex]Volume \ of \ the \ cone \approx \frac{1}{3} \times 3.14 \times 5^2 \times 8\approx \mathbf{209.3\, u^3}[/tex]
209.3 u³ is equivalent to H
83 961 392 7542 gives; IT HAS THE EDGE
Learn more about calculating the volume of a cone here:
https://brainly.com/question/3152808
